10806-Dijkstra, Dijkstra.

// The Problem of the minimum cost and the maximum flow (it refers to the problem that the total cost is the minimum when the total traffic is the maximum)

// Because you do not need to find the total stream for this question, it is more suitable to use Edmonds-Karp than pre-stream propulsion.

// Note that cost [u] [PA [u] =-cost [PA [u] [u]; in this way, we can ensure that when the stream is resized again, we can "TOP" the stream that was not suitable before.

// Do not use dijisktra twice, for example, the following data

6. <br/> 8 <br/> 1 2 1 <br/> 2 3 1 <br/> 3 6 1 <br/> 1 4 100 <br/> 4 5 100 <br/> 5 6 100 <br/> 1 3 10 <br/> 2 6 10 </P> <p>

# Include <iostream> <br/> # include <cstdio> <br/> # include <queue> <br/> # include <memory. h> </P> <p> using namespace STD; <br/> const int max = 105, INF = 1 <30; <br/> bool INQ [Max]; <br/> int m, n, Cap [Max] [Max], flow [Max] [Max], cost [Max] [Max], <br/> d [Max], a [Max], pa [Max]; <br/> queue <int> q; <br/> int BFS () {<br/> int f = 0, c = 0, U, V; <br/> memset (flow, 0, sizeof (flow )); <br/> while (1) {<br/> memset (INQ, 0, sizeof (INQ); <B R/> for (INT I = 2; I <= N; I ++) <br/> d [I] = inf; <br/> d [1] = 0; <br/> INQ [1] = true; <br/> q. push (1); <br/> while (! Q. empty () {<br/> U = Q. front (); q. pop (); <br/> INQ [u] = false; <br/> for (V = 1; v <= N; V ++) {<br/> If (u = 5 & V = 2) {<br/>/* cout <Endl <CAP [u] [v] <"<br/> <flow [u] [v] <Endl; <br/> cout <D [v] <"" <D [u] <Endl; <br/> cout <cost [u] [v] <Endl; */<br/>}< br/> If (Cap [u] [v]> flow [u] [v] & D [v]> d [u] + cost [u] [v]) {<br/> d [v] = d [u] + cost [u] [v]; <br/> pa [v] = u; <br/> If (! INQ [v]) {<br/> INQ [v] = true; <br/> q. push (V); <br/>}< br/> If (d [N] = inf) <br/> break; <br/> int A = inf; <br/> for (u = N; u! = 1; u = pa [u]) A = A <(Cap [PA [u] [u]-flow [PA [u] [u])? <Br/> A: CAP [PA [u] [u]-flow [PA [u] [u]); <br/> for (u = N; u! = 1; u = pa [u]) {<br/> flow [PA [u] [u] + =; <br/> flow [u] [PA [u]-=; <br/> cost [u] [PA [u] =-cost [PA [u] [u]; <br/>}< br/> // cout <"A:" <A <"D:" <D [N] <Endl; <br/> U = N; <br/>/* While (u! = 1) {<br/> cout <u <"<-- (" <D [PA [u] <")" <"--"; <br/> U = pa [u]; <br/>}< br/> cout <1 <Endl; */<br/> C + = d [N] * A; <br/> F + = A; <br/> If (F = 2) {<br/> return C; <br/>}< br/> return 0; <br/>}< br/> int main () <br/>{< br/> freopen ("I .txt", "r", stdin); <br/> int U, V, C, ans; <br/> while (CIN> N, N) {<br/> CIN> m; <br/> memset (Cap, 0, sizeof (CAP )); <br/> memset (cost, 0, sizeof (cost); <br/> for (INT I = 0; I <m; I ++) {<br/> CIN> U> V> C; <br/> CAP [u] [v] = CAP [v] [u] = 1; <br/> cost [u] [v] = cost [v] [u] = C; <br/> // CAP [u] [v] = 1; <br/> // cost [u] [v] = C; <br/>}< br/> ans = BFS (); <br/> If (ANS) <br/> cout <ans <Endl; <br/> else <br/> cout <"back to jail" <Endl; <br/>}< br/> return 0; <br/>}< br/>